Problem: $\begin{aligned} &f(t)=\dfrac{2t+7}{t-3} \\\\ &h(n)=-2n^2-3n+1 \end{aligned}$ $f(h(-1))=$
Solution: When evaluating composite functions, we work our way inside out. To evaluate $f(h(-1))$, let's first evaluate $h(-1)$. Then we'll plug that result into $f$ to find our answer. Let's evaluate $h({-1})$. $\begin{aligned}h(n)&=-2n^2-3n+1\\\\ h({-1})&=-2({-1})^2-3({-1})+1 ~~~~~~~~~~\text{Plug in }n={-1}\\\\ &=-2(1)+3+1\\\\ &={2}\end{aligned}$ We now know that $f(h({-1}))$ is the same as $f({2})$ because $h({-1}) = {2}$. Let's evaluate $f({2})$. $\begin{aligned}f(t)&=\dfrac{2t+7}{t-3}\\\\ f({{2}})&=\dfrac{2({2})+7}{({2})-3}~~~~~~~~~~\text{Plug in }t={2}\\\\ &=\dfrac{11}{-1}\\\\ &=-11\\\\\end{aligned}$ The answer: $f(h(-1))= -11$